### Solve the quadratic equation x^{2
}- 5x + 6 = 0 to get x=5-6/x. Iterate the
function 5-6/x and graph the starting number vs the limit of the infinite
sequence that is formed

**2b.**We'll iterate the function
^{}starting with different numbers, then graph
the limits of the infinite sequences we get.

Starting number 1. Putting 1->x, we get 5-(6/1)=^{-}1. Putting this output
number ^{-}1 in for x now, we get 5-(6/-1) = 11. Putting 11 -> x we get
5-(6/11)= 4.45..

So we get this infinite sequence:

1, -1, 11,
4.45.., 3.653..., 3.3575..., 3.2129..., ...which goes to 3 as the
limit.

Starting no. -100, 5.06, 3.81422..., 3.4269...,
3.2491...,3.1533..., ...which goes to 3 as the limit.

Notice that starting with 2 we get a constant sequence 2, 2, 2, ... On
the graph 2 goes to 2.

All other starting numbers lead to an infinite
sequence which goes to 3 as the limit.., **except an infinite number of
starting numbers like zero, and 6/5 that at some point make the denominator
go to
0 and make the fraction 'blow up' and thus has no answer. These numbers
put a hole in the graph. **Can you find a rule for these numbers that
make the denominato 0? What is the limit of the sequence of these numbers?

To order Don's
materials

Back to Ch. 8, part 2- iteration

Mathman home